Search found 179 matches
- Wed Aug 01, 2018 9:40 am
- Forum: Analysis
- Topic: Multiplicity of root
- Replies: 1
- Views: 5228
Re: Multiplicity of root
Given the function $f(x)=e^x-x-1$ prove that $0$ is a zero of $f$ of multiplicity $2$. It suffices to prove that the limit $\displaystyle \lim \limits_{x \rightarrow 0} \frac{f(x)}{x^2}$ is finite. However, \begin{align*} \lim_{x\rightarrow 0} \frac{f(x)}{x^2} &= \lim_{x\rightarrow 0} \frac{e^x...
- Wed Aug 01, 2018 9:35 am
- Forum: General Mathematics
- Topic: A sum!
- Replies: 4
- Views: 8139
Re: A sum!
Tolaso J Kos wrote:Evaluate the following sum:
$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}}$$
Source
$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}} = \frac{3\sqrt{7}}{112} $$
Full solution tomorrow morning !
- Wed Aug 01, 2018 9:33 am
- Forum: Number theory
- Topic: Irrational number
- Replies: 1
- Views: 10237
Re: Irrational number
Let $p \leq N$ be the last prime. If we prove that between $p$ and $N$ does not exist a number that has $p$ as a factor we are done. So, we need to prove that $2p>N$. But this is exactly what Bertrand's postulate says.
- Tue Jun 19, 2018 7:20 pm
- Forum: Calculus
- Topic: A definite Integral
- Replies: 2
- Views: 4390
Re: A definite Integral
Are you sure about the upper limit? Should not it be $\frac{\pi}{2}$ ?
- Sat May 26, 2018 1:31 pm
- Forum: Real Analysis
- Topic: A zeta limit
- Replies: 2
- Views: 6276
Re: A zeta limit
Using the above fact we get that $\zeta^{(n)}(0) \sim -n!$.
- Sat May 26, 2018 1:28 pm
- Forum: Real Analysis
- Topic: A zeta limit
- Replies: 2
- Views: 6276
Re: A zeta limit
We know that the function $\displaystyle f(z) \equiv \zeta(z) + \frac{1}{1-z}$ is a holomorphic function. The Taylor series around $0$ is $$\zeta(z) + \frac{1}{1-z} = \sum_{n=0}^{\infty} \left( \frac{\zeta^{(n)}(0)}{n!} + 1 \right) z^n$$ which converges forall $z \in \mathbb{C}$ thus $\displaystyle ...
- Sat May 26, 2018 1:18 pm
- Forum: Real Analysis
- Topic: A limit with Euler's totient function
- Replies: 1
- Views: 5229
Re: A limit with Euler's totient function
We are quoting a theorem by Omran Kouba: Theorem: Let $\alpha$ be a positive real number and let $\{a_n\}_{n \in \mathbb{N}}$ be a sequence of positive real numbers such that $$\lim_{n \rightarrow +\infty} \frac{1}{n^\alpha} \sum_{k=1}^{n} a_k = \ell$$ For every continuous function $f$ on the interv...
- Sat May 26, 2018 1:11 pm
- Forum: Number theory
- Topic: Series with least common multiple.
- Replies: 1
- Views: 5228
Re: Series with least common multiple.
An elementary approach would be as follows: For all $M \in \mathbb{N}^+$ of the form $M=p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ the number of solutions $$\mathrm{lcm} (m, n) =M$$ is given by $(2a_1+1) (2a_2+1) \cdots (2a_k+1)$. It follows that the given series equals $$\sum_{M =1}^{\infty} \frac{1}{M^...
- Sat May 26, 2018 1:02 pm
- Forum: Real Analysis
- Topic: On an evaluation of an arctan limit
- Replies: 1
- Views: 5406
Re: On an evaluation of an arctan limit
Why not prove the more general result? Let $f:[0, 1] \rightarrow (0, +\infty)$ be a bounded integrable function. Then: \[\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \frac{f\left ( \frac{k}{n} \right )}{1+2\sqrt{\frac{1}{n} f\left ( \frac{k}{n} \right )+1}} = \frac{1}{3} \int_{0}^{1} f(x)...
- Sat May 26, 2018 1:00 pm
- Forum: Competitions
- Topic: 16 th Cuban Mathematical Competition of Universities [Problem 5]
- Replies: 1
- Views: 6059
Re: 16 th Cuban Mathematical Competition of Universities [Problem 5]
Greetings, We are focusing on the $\alpha$' s lying in the interval $(0, 1)$. That is because each term of the series is $1$ periodic. Let $\mathbb{Z} \ni k >0$ and let $n$ be the maximal integer for which it holds \[k-1 <n\alpha < k\] Since it holds that $\left \{ n \alpha \right \} \geq 1-\alpha$ ...